I'm sorry for a dumb problem in advance but the problem states
There exist two ordered triplets $(a_1, b_1 ,c_1)$ and $(a_2 , b_2 , c_2)$ for $(a, b, c)$ for which the equation $4x^2 - 4xy + ay^2 + bx + cy + 1 = 0$ represents a pair of identical straight lines in $x-y$ plane. Find the value of $a_1+ b_1 +c_1 +a_2 + b_2 +c_2$
The first thing I tried was $h^2-4ab=0$ and by that, I got the value of $a=1$ but I cannot seem to move further and find values of $b$ and $c$
The author just simply writes
By observation, directly equation must be $(4x^2-4xy+y^2)+2(2x-y)+1=0$ or $(4x^2-4xy+y^2)-2(2x-y)+1=0$ and gives the answer $2$
PS: it's a highschool level problem.
The equation of a pair of identical straight lines is $$(ux+vy+w)^2=0$$ Where $u,v,w$ are constants. We can match coefficients to see $$u^2=4$$ $$2uv=-4$$ $$w^2=1$$ WLOG say $u=2$. This gives $v=-1$. Now we can case on whether $w=1$ or $w=-1$. Expanding the squared trinomial in both cases should give us the values for $a,b,c$.
Instead of directly solving for $a,b,c$, we can get the sum indirectly. We know that the factored form looks like $$(2x-y\pm 1)^2=0$$ Note that the LHS expands into $$4x^2-4xy+ay^2+bx+cy+1$$ If we plug in $x=1,y=1$ into the expression we get $a+b+c+1$. If we plug in $x=1,y=1$ into our factored form we get $$(1\pm 1)^2$$ Which has value $0$ or $4$. Hence, the total sum of all possible values is $(0-1)+(4-1)=\boxed{2}$.